Test bank basic statistics for business and economics 4th ce

F o r O r d e r T h i s A n d A n y O t h e r T e s t

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2.

3.

It is based on two sets of degrees of freedom

awa rd:

It is positively skewed.

1

→

0 .

As the values of X increase, the F curve approaches the X-axis and eventually equals zero.

It is asymptotic.

0 0 p

Multiple Choice

Difficulty: Easy

Which of the following is NOT a characteristic of ...

Learning Objective: 1

o i n t s

Wh ich of the foll owi ng is NO Ta cha rac teri stic of . ..

Whi ch of the follo wing is NOT a char acte ristic of the F distri buti on?

award:

10.00 points

i. One characteristic of the F distribution is tha... i. One characteristic of the F distribution is that F cannot be negative.T2 ii. One characteristic of the F distribution is that computed F can only range between -1 and +1.F2 iii. The shape of the F distribution is positively skewed (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii).

→

(i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. One characteristic of the F distribution is tha...

Learning Objective: 1

award:

10.00 points

i. One characteristic of the F distribution is tha... i. One characteristic of the F distribution is that F cannot be negative. ii. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. iii. The F distribution's curve is positively skewed.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. One characteristic of the F distribution is tha...

Learning Objective: 1

1 of 33

4.

award:

10.00 points

i. One characteristic of the F distribution is tha... i. One characteristic of the F distribution is that computed F can only range between -1 and +1. ii. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. iii. The F distribution's curve is positively skewed. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(ii) and (iii) are correct statements but not (i).

All statements are false

5.

Multiple Choice

Difficulty: Hard

i. One characteristic of the F distribution is tha...

Learning Objective: 1

award:

10.00 points

i. The F distribution is positively skewed and its... i. The F distribution is positively skewed and its values may range from 0 to plus infinity. ii. The F distribution's curve is positively skewed. iii. There is one, unique F distribution for a F-statistic with 29 degrees of freedom in the numerator and 28 degrees of freedom in the denominator.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

6.

Multiple Choice

Difficulty: Hard

i. The F distribution is positively skewed and its...

Learning Objective: 1

award:

10.00 points

i. The shape of the F distribution is determined b... i. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. ii. The F distribution's curve is positively skewed. iii. Unlike Student's t distribution, there is only one F distribution. (i), (ii) and (iii) are all correct statements

→

(i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

7.

Multiple Choice

Difficulty: Hard

i. The shape of the F distribution is determined b...

Learning Objective: 1

award:

10.00 points

i. The F distribution is positively skewed and its... i. The F distribution is positively skewed and its values may range from 0 to plus infinity. ii. The F distribution's curve is positively skewed iii. Like Student's t distribution, a change in the degrees of freedom causes a change in the shape of the F distribution.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The F distribution is positively skewed and its...

Learning Objective: 1

2 of 33

award:

8.

10.00

points

i. The test statistic used in ANOVA is Fii. The ca... i. The test statistic used in ANOVA is F ii. The calculated F value must be equal to or greater than zero (0) iii. The shape of the F distribution is symmetrical (i), (ii) and (iii) are all correct statements

→

(i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The test statistic used in ANOVA is Fii. The ca...

Learning Objective: 1

award:

9.

10.00

points

i. The shape of the F distribution is determined b... i. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. ii. Like Student's t distribution, a change in the degrees of freedom causes a change in the shape of the F distribution. iii. The calculated F value must be equal to or greater than zero (0)

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The shape of the F distribution is determined b...

Learning Objective: 1

award:

10.

10

points

.00

i. The F distribution is positively skewed and its... i. The F distribution is positively skewed and its values may range from 0 to plus infinity. ii. The F distribution's curve is positively symmetrical. iii. Like Student's t distribution, a change in the degrees of freedom causes a change in the shape of the F distribution. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii).

→

(i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The F distribution is positively skewed and its...

Learning Objective: 1

award:

11.

10.00

points

i. One characteristic of the F distribution is tha... i. One characteristic of the F distribution is that computed F can only range between -1 and +1. ii. The F distribution's curve is positively skewed. iii. Like Student's t distribution, a change in the degrees of freedom causes a change in the shape of the F distribution. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. One characteristic of the F distribution is tha...

Learning Objective: 1

3 of 33

award:

12.

10

points

.00

i. The test statistic used in ANOVA is tii. The ca... i. The test statistic used in ANOVA is t ii. The calculated F value must be equal to or greater than one (1) iii. The shape of the F distribution is symmetrical. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

→

All statements are false

Multiple Choice

Difficulty: Hard

i. The test statistic used in ANOVA is tii. The ca...

Learning Objective: 1

award:

13.

10

points

.00

What distribution does the F distribution approach...

What distribution does the F distribution approach as the sample size increases?

Binomial

→

Normal Poisson Exponential None of these is correct

Multiple Choice

Difficulty: Medium

What distribution does the F distribution approach...

Learning Objective: 1

14. 10.00 award:

points

An F statistic is: An F statistic is: a ratio of two means.

→

a ratio of two variances. the difference between three means. a population parameter.

Multiple Choice

Difficulty: Easy

An F statistic is:

Learning Objective: 1

award:

15.

10

points

.00

Which statement is correct about the F distributio...

Which statement is correct about the F distribution?

→

Cannot be negative Cannot be positive Is the same as the t distribution Is the same as the z distribution None of these is correct

Multiple Choice

Difficulty: Medium

Which statement is correct about the F distributio...

Learning Objective: 1

4 of 33

award:

16.

10

points

.00

Which of the following are characteristics of the ...

Which of the following are characteristics of the F distribution?

There is a "family" of F distributions The F distribution is continuous.

The F distribution cannot be negative.

→

All of the above are characteristics of the F distribution.

Multiple Choice

Difficulty: Medium

Which of the following are characteristics of the ...

Learning Objective: 1

award:

17.

10

points

.00

i. If the computed value of F is 0.99 and the crit... i. If the computed value of F is 0.99 and the critical value is 3.89, we would not reject the null hypothesis. ii. When comparing two population variances we use the F distribution. iii. A one way ANOVA is use to compare several treatment means.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. If the computed value of F is 0.99 and the crit...

Learning Objective: 2

award:

18.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

Using Excel to assist in the comparison, what test would be used?

ANOVA: Single Factor ANOVA: Two-Factor with Replication F-Test Two Sample for Variances t-Test: Paired Two Sample for Means

→

We need the raw data in order to use the F-test in Excel

Multiple Choice

Difficulty: Medium

Two accounting professors decided to compare the v...

Learning Objective: 2

5 of 33

19.

award:

10.00 points

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What is H0?

→

2 2 σ 1=σ 2 2 2 σ 1≠σ 2

µ1 = µ2 µ1 ≠ µ2 None of these statements are true Multiple Choice Two accounting professors decided to compare the v...

20.

Difficulty: Easy Learning Objective: 2

award:

10.00 points

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What is H1?

→

2 2 σ 1=σ 2 2 2 σ 1≠σ 2

µ1 = µ2 µ1 ≠ µ2 None of these statements are true

Multiple Choice Two accounting professors decided to compare the v...

21.

Difficulty: Easy Learning Objective: 2

award:

10.00 points

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What are the degrees of freedom for the numerator of the F ratio?

8

→

9 10 18 20

Multiple Choice Two accounting professors decided to compare the v...

Difficulty: Medium Learning Objective: 2

6 of 33

award:

22.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What are the degrees of freedom for the denominator of the F ratio?

20 18 10

→

9 8

Multiple Choice Two accounting professors decided to compare the v...

Difficulty: Medium Learning Objective: 2

award:

23.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What is the critical value of F at the 0.01 level of significance?

5.85

→

5.35 6.51 4.03 None of these statements are true

Multiple Choice Two accounting professors decided to compare the v...

Difficulty: Medium Learning Objective: 2

award:

24.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

What is the critical value of F at the 0.05 level of significance?

5.85 5.35

→

3.18 4.03 None of these statements are true

Multiple Choice Two accounting professors decided to compare the v...

Difficulty: Medium Learning Objective: 2

7 of 33

award:

25.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

The calculated F ratio is:

→

3.484 1.867 3.18 5.35 None of these statements are true

Multiple Choice Two accounting professors decided to compare the v...

Difficulty: Hard Learning Objective: 2

award:

26.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

At the 1% level of significance, what is the decision?

Reject the null hypothesis and conclude the variance is different.

Fail to reject the null hypothesis and conclude the variance is different.

Reject the null hypothesis and conclude the variance is the same.

→

Fail to reject the null hypothesis and conclude the variance is the same.

None of these statements are true

Multiple Choice

Difficulty: Hard

Two accounting professors decided to compare the Learning Objective: 2 v...

award:

27.

10

points

.00

Two accounting professors decided to compare the v...

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:

At the 5% level of significance, what is the decision?

→

Reject the null hypothesis and conclude the variance is different.

Fail to reject the null hypothesis and conclude no significant difference in the variance.

Reject the null hypothesis and conclude the variance is the same.

Fail to reject the null hypothesis and conclude the variance is the same.

None of these statements are true

Multiple Choice

Difficulty: Hard

Two accounting professors decided to compare the Learning Objective: 2 v...

8 of 33

award:

28.

10

points

.00

Analysis of variance is used to: Analysis of variance is used to: compare nominal data. compute t test. compare population proportion.

→

simultaneously compare several population means.

None of these statements are true.

Multiple Choice

Difficulty: Medium

Analysis of variance is used to:

Learning Objective: 3

award:

29.

10

points

.00

i. The statistical technique used to test the equa... i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA) ii. To employ ANOVA, the populations should have approximately equal standard deviations. iii. To employ ANOVA, the populations being studied must be approximately normally distributed.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The statistical technique used to test the equa...

Learning Objective: 3

award:

30.

10

points

.00

i. The statistical technique used to test the equa... i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA) ii. To employ ANOVA, the populations should have approximately equal standard deviations. iii. The least number of sources of variation in ANOVA is two.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The statistical technique used to test the equa...

Learning Objective: 3

award:

31.

10

points

.00

i. The statistical technique used to test the equa... i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA) ii. To employ ANOVA, the populations need not have equal standard deviations. iii. To employ ANOVA, the populations being studied must be approximately normally distributed. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii).

→

(i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The statistical technique used to test the equa...

Learning Objective: 3

9 of 33

award:

32.

10

points

.00

i. To employ ANOVA, the populations should have ap... i. To employ ANOVA, the populations should have approximately equal standard deviations. ii. To employ ANOVA, the populations being studied must be approximately normally distributed. iii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of variance (ANOVA)

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. To employ ANOVA, the populations should have

Learning Objective: 3

ap...

award:

33.

10

points

.00

i. To employ ANOVA, the populations need not have ... i. To employ ANOVA, the populations need not have equal standard deviations. ii. To employ ANOVA, the populations being studied need not be normally distributed. iii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of deviation. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

→

All statements are false

Multiple Choice i. To employ ANOVA, the populations need not have ...

Difficulty: Hard Learning Objective: 3

award:

34.

10

points

.00

i. To employ ANOVA, the populations being studied ... i. To employ ANOVA, the populations being studied must be approximately normally distributed. ii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of variance (ANOVA) iii. The least number of sources of variation in ANOVA is two.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. To employ ANOVA, the populations being studied ...

Learning Objective: 3

10 of 33

award:

35.

10

points

.00

i. To employ ANOVA, the populations should have ap... i. To employ ANOVA, the populations should have approximately equal standard deviations. ii. To employ ANOVA, the populations being studied must be approximately normally distributed. iii. The least number of sources of variation in ANOVA is two.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice i. To employ ANOVA, the populations should have ap...

Difficulty: Hard Learning Objective: 3

award:

36.

10

points

.00

i. The rejection region for analysis of variance i... i. The rejection region for analysis of variance is in the upper tail of the F distribution ii. In ANOVA, k -1 degrees of freedom are associated with the numerator of the F ratio. iii. In ANOVA, k-1 degrees of freedom are associated with the denominator of the F ratio. (i), (ii) and (iii) are all correct statements

→

(i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The rejection region for analysis of variance i...

Learning Objective: 3

award:

37.

10

points

.00

Which one of the following is not assumed in the u...

Which one of the following is not assumed in the use of ANOVA? The populations follow a normal distribution.

→

The samples have equal standard deviations.

The populations have equal standard deviations.

The populations are independent.

Multiple Choice

Difficulty: Easy

Which one of the following is not assumed in the u...

Learning Objective: 3

award:

38.

10

points

.00

The F test statistic is the ratio of the The F test statistic is the ratio of the Estimate of the population mean based on the differences among the sample standard deviations to the estimate of the population variance based on the variation within the samples.

→

Estimate of the population variance based on the differences among the sample means to the estimate of the population variance based on the variation within the samples.

Estimate of the population variance based on the sums of the sample means to the estimate of the population variance based on the variation within the samples.

None of the above.

Multiple Choice

Difficulty: Easy

The F test statistic is the ratio of the

Learning Objective: 3

11 of 33

award:

39.

10

points

.00

A random sample of 30 executives from companies wi...

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The following table summarized the results:

Using this output, what conclusions can you draw?

Since the calculated F-value is smaller than the F-critical value, there is a significant difference in the incomes of these 3 groups Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the incomes of these three groups Since the P-value is 0.001, there is an 10% chance of these results happening when there is no significant difference in the incomes of these three groups

→

Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that the three groups with different levels of education do not all have the same incomes Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that all three groups with different levels of education have the same incomes

Multiple Choice

Difficulty: Medium

A random sample of 30 executives from companies wi...

Learning Objective: 4

40. 10.00 award:

points

A random sample of 30 executives from companies wi...

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The following table summarized the results:

Using this output, what conclusions can you draw?

Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the incomes of these three groups Since the calculated F-value is smaller than the F-critical value, there is a significant difference in the incomes of these 3 groups Since the _P-value is 0.001, there is an 10% chance of these results happening when there is no significant difference in the incomes of these three groups Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that all three groups with different levels of education have the same incomes

→

Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that the three groups with different levels of education do not all have the same incomes

Multiple Choice A random sample of 30 executives from companies wi...

Difficulty: Medium Learning Objective: 4

12 of 33

41.

award:

10.00 points

A large department store examined a sample of the ...

A large department store examined a sample of the 18 credit card sales and recorded the amounts charged for each of three types of credit cards: MasterCard, Visa and American Express. Six MasterCard sales, seven Visa and five American Express sales were recorded. The store used ANOVA to test if the mean sales for each credit card were equal. What are the degrees of freedom for the F statistic?

18 in the numerator, 3 in the denominator 3 in the numerator, 18 in the denominator

→

2 in the numerator, 15 in the denominator 0 in the numerator, 15 in the denominator None of these statements are true

42.

Multiple Choice

Difficulty: Medium

A large department store examined a sample of the ...

Learning Objective: 4

award:

10.00 points

Suppose that an automobile manufacturer designed a...

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline to use. The four grades are: below regular, regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. Assuming any grade can be used at the 0.05 level, what is the critical value of F using 0.05 level of significance?

1.96

→

4.07 2.33 12.00 None of these statements are true

Multiple Choice Suppose that an automobile manufacturer designed a...

43.

Difficulty: Medium Learning Objective: 4

award:

10.00 points

In ANOVA, an F statistic is used to test a null hy...

In ANOVA, an F statistic is used to test a null hypothesis such as: 2 2 2 Ho : σ 1 = σ 2 = σ 3 2 2 2 Ho: σ 1 ≠ σ 2≠ σ 3

→

Ho: µ1 = µ2= µ3 Ho: µ1 ≠ µ2≠ µ2 None of these statements are true

Multiple Choice

Difficulty: Easy

In ANOVA, an F statistic is used to test a null hy...

Learning Objective: 4

13 of 33

44.

award:

10.00 points

Three different advertisements were used to sell a...

Three different advertisements were used to sell a popular toy. In computing F, how many degrees of freedom are there in the numerator?

0 1

→

2 3 None of these statements are true

45.

Multiple Choice

Difficulty: Medium

Three different advertisements were used to sell a...

Learning Objective: 4

award:

10.00 points

Suppose a package delivery company purchased 14 tr...

Suppose a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from B and five from manufacturer C. The cost of maintaining each truck was recorded. The company used ANOVA to test if the mean maintenance cost for trucks from each manufacturer were equal. To apply the F test, how many degrees of freedom are in the denominator?

2 3

→

11 14 None of these statements are true

Multiple Choice Suppose a package delivery company purchased 14 tr...

46.

Difficulty: Medium Learning Objective: 4

award:

10.00 points

In an effort to determine the most effective way t...

In an effort to determine the most effective way to teach safety principles to a group of employees, four different methods were tried. Some employees were given programmed instruction booklets and worked through the course at their own pace. Other employees attended lectures. A third group watched a television presentation, and a fourth group was divided into small discussion groups. A high of 10 was possible. Samples of five tests were selected from each group. The test grade results were:

At the 0.01 level, what is the critical value?

1.00 1.96 3.24

→

5.29 None of these statements are true

Multiple Choice

Difficulty: Medium

In an effort to determine the most effective way t...

Learning Objective: 4

14 of 33

award:

47.

10

points

.00

In an effort to determine the most effective way t...

In an effort to determine the most effective way to teach safety principles to a group of employees, four different methods were tried. Some employees were given programmed instruction booklets and worked through the course at their own pace. Other employees attended lectures. A third group watched a television presentation, and a fourth group was divided into small discussion groups. A high of 10 was possible. Samples of five tests were selected from each group. The test grade results were:

Excel's summary results at the 0.05 level produce the following output:

Using this output, what conclusions can you draw?

Since the calculated F-value is smaller than the F-critical value, there is a significant difference in the methods of teaching Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the methods of teaching Since the P-value is 0.001, there is an 10% chance of these results happening when there is no significant difference in the methods of teaching

→

Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that all teaching methods do not give equal test results Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that all teaching methods give equal test results

Multiple Choice

Difficulty: Medium

In an effort to determine the most effective way t...

Learning Objective: 4

15 of 33

award:

48.

10

points

.00

Random sample of executives from companies with as...

Random sample of executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The following MegaStat output summarized the results:

Using this output, what conclusions can you draw?

The total size of the sample used was 24 Since the P-value is 0.0008, there is an 8% chance of these results happening when there is no significant difference in the incomes of these three groups

→

Since the calculated F-value is large, and the P-value is so small, there is strong evidence to suggest that the three groups with different levels of education do not all have the same incomes Since the calculated F-value is large, and the P-value is so small, there is strong evidence to suggest that all three groups with different levels of education have the same incomes A & D are true

Multiple Choice

Difficulty: Medium

Random sample of executives from companies with as...

Learning Objective: 4

49. 10.00 award:

points

A random sample of 25 executives from companies wi...

A random sample of 25 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The following Excel printout summarized the results:

Since the calculated F-value is smaller than the F-critical value, there is a significant difference in the incomes of these 3 groups Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the incomes of these three groups Since the P-value is 0.001, there is an 10% chance of these results happening when there is no significant difference in the incomes of these three groups Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that all three groups with different levels of education have the same incomes

→

Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that the three groups with different levels of education do not all have the same incomes

Multiple Choice

Difficulty: Medium

A random sample of 25 executives from companies wi...

Learning Objective: 4

16 of 33

award:

50.

10

points

.00

Suppose that an automobile manufacturer designed a...

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline to use. The four grades are: below regular, regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. Is there a difference in the performance between the four grades of gas?

Using the printout given by Excel, what to you conclude?

Since the calculated F-value is smaller than the F-critical value, there is a difference in the performance of these four grades of gas Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the performance of these four grades of gas Since the P-value is large at 0.899, there is an 89% chance of these results happening when there is no significant difference in the performance in the four grades of gas

→

B & C are true None of these statements are true

Multiple Choice Suppose that an automobile manufacturer designed a...

Difficulty: Medium Learning Objective: 4

17 of 33

award:

51.

10

points

.00

The plant manager believes that the temperature in...

The plant manager believes that the temperature in the packaging area of the plant affects the daily rate of production. To investigate, the plant temperature is set at 18 degrees, 20 degrees, and 22 degrees. The number of units produced at each of these temperatures for a sample of days is collected.

Using this output, what conclusions can you draw?

The total size of the sample used was 24 Since the P-value is 0.0024, there is a 2% chance of these results happening when there is no significant difference in the number of units produced at the three temperatures Since the calculated F-value is large, and the P-value is so small, there is strong evidence to suggest that the number of units produced at the three temperatures is all the same

→

Since the calculated F-value is large, and the P-value is so small, there is strong evidence to suggest that the number of units produced at the three temperatures is not all the same A & D are true

Multiple Choice

Difficulty: Medium

The plant manager believes that the temperature in...

Learning Objective: 4

award:

52.

10

points

.00

If an ANOVA test is conducted and the null hypothe...

If an ANOVA test is conducted and the null hypothesis is rejected, what does this indicate?

Too many degrees of freedom No difference between the population means

→

A difference between at least one pair of population means None of these statements are true

Multiple Choice

Difficulty: Medium

If an ANOVA test is conducted and the null hypothe...

Learning Objective: 4

18 of 33

award:

53.

10

points

.00

Several employees have submitted different methods...

Several employees have submitted different methods of assembling a product. Sample data for each method are:

How many treatments are there?

3

→

4 12 0 None of these statements are true

Multiple Choice

Difficulty: Medium

Several employees have submitted different methods...

Learning Objective: 4

award:

54.

10

points

.00

In an effort to determine the most effective way t...

In an effort to determine the most effective way to teach safety principles to a group of employees, four different methods were tried. Some employees were given programmed instruction booklets and worked through the course at their own pace. Other employees attended lectures. A third group watched a television presentation, and a fourth group was divided into small discussion groups. A high of 10 was possible. Samples of five tests were selected from each group. The test grade results were:

Using Excel to assist in determining if all methods generate the same results, what test would be used?

→

ANOVA: Single Factor ANOVA: Two-Factor with Replication ANOVA: Two-Factor without Replication F-Test Two Sample for Variances

t-

Test: Paired Two Sample for Means Multiple Choice

Difficulty: Medium

In an effort to determine the most effective way t...

Learning Objective: 4

19 of 33

award:

55.

10

points

.00

Suppose that an automobile manufacturer designed a...

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline to use. The four grades are: below regular, regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. Is there a difference in the performance between the four grades of gas?

If you were to use Excel to assist in your solution to this problem, which test would you use?

→

ANOVA: Single Factor ANOVA: Two-Factor with Replication ANOVA: Two-Factor without Replication F-Test Two Sample for Variances

t-

Test: Paired Two Sample for Means Multiple Choice

Difficulty: Medium

Suppose that an automobile manufacturer designed a...

Learning Objective: 4

award:

56.

10

points

.00

An electronics company wants to compare the qualit...

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. If ANOVA is used to compare the average number of defects, the treatments would be defined as: the number of cell phones sampled. the average number of defects.

The total number of phones

→

The four companies.

Multiple Choice

Difficulty: Easy

An electronics company wants to compare the

Learning Objective: 4

qualit...

award:

57.

10

points

.00

An electronics company wants to compare the qualit...

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. Using Excel, what test is used to compare the average number of defects?

→

ANOVA: Single Factor ANOVA: Two-Factor with Replication ANOVA: Two-Factor without Replication F-Test Two Sample for Variances

t-

Test: Paired Two Sample for Means Multiple Choice An electronics company wants to compare the qualit...

Difficulty: Medium Learning Objective: 4

20 of 33

58.

award:

10.00 points

Suppose a package delivery company purchased 14 tr...

Suppose a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from B and five from manufacturer C. The cost of maintaining each truck was recorded. The company wants to test if the mean maintenance cost for trucks from each manufacturer were equal. Using Excel, what test would be used?

→

ANOVA: Single Factor ANOVA: Two-Factor with Replication ANOVA: Two-Factor without Replication F-Test Two Sample for Variances

t-

Test: Paired Two Sample for Means Multiple Choice Suppose a package delivery company purchased 14 tr...

59.

Difficulty: Medium Learning Objective: 4

award:

10.00 points

A preliminary study of hourly wages paid to unskil...

A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?

Mean hourly wages of unskilled employees all areas are equal

→

Mean hourly wages in at least 2 metropolitan areas are different More degrees of freedom are needed None of these is correct

60.

Multiple Choice

Difficulty: Medium

A preliminary study of hourly wages paid to unskil...

Learning Objective: 4

award:

10.00 points

Using the Excel printout below to compare the mean...

Using the Excel printout below to compare the mean annual incomes for executives with Undergraduate and Master's Degree or more, the following statements can be made:

Since the calculated F-value is smaller than the F-critical value, there is no significant difference in the incomes of these two groups Since the calculated F-value is smaller than the F-critical value, there is a significant difference in the incomes of these two groups Since the P-value is 0.62, there is a 62% chance of these results happening when there is no significant difference in the incomes of these two groups Since the calculated F-value is larger than the F-critical value, and the P-value is so small, there is strong evidence to suggest that both groups with different levels of education have the same incomes

→

A & C are true

Multiple Choice

Difficulty: Medium

Using the Excel printout below to compare the mean...

Learning Objective: 4

21 of 33

award:

61.

10

points

.00

Using the MegaStat printout below to compare the m...

Using the MegaStat printout below to compare the mean annual incomes for executives with Undergraduate and High School or less, the following statements can be made:

Since the 95% confidence interval does not contain the value 0, there is no significant difference in the incomes of these two groups Since the 95% confidence interval does not contain the value 0, there is a significant difference in the incomes of these two groups Since the P-value is small, there is a strong chance of these results happening when there is no significant difference in the incomes of these two groups Since the P-value is small, there is very little chance of these results happening when there is no significant difference in the incomes of these two groupstherefore those executives with an undergraduate degree make more than their counterparts without

→

B & D are true

Multiple Choice Using the MegaStat printout below to compare the m...

Difficulty: Medium Learning Objective: 4

award:

62.

10

points

.00

i. If the computed value of F is 4.01 and the crit... i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal.F5 ii. Rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA is μ1 - μ2 ≠ μ3 (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

→

All statements are false

Multiple Choice

Difficulty: Hard

i. If the computed value of F is 4.01 and the crit...

Learning Objective: 4

22 of 33

award:

63.

10

points

.00

i. The alternate hypothesis for ANOVA states that ... i. The alternate hypothesis for ANOVA states that not all the means are equal. ii. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly. iii. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. (i), (ii) and (iii) are all correct statements

→

(i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The alternate hypothesis for ANOVA states that ...

Learning Objective: 4

award:

64.

10

points

.00

i. For the population means, the alternate hypothe... i. For the population means, the alternate hypothesis used in the analysis of variance test states that µ1 = µ2 = µ3. ii. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly. iii. The null hypothesis for an ANOVA is µ1 - µ2 = µ3 (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. For the population means, the alternate hypothe...

Learning Objective: 4

award:

65.

10

points

.00

i. For the population means, the alternate hypothe... i. For the population means, the alternate hypothesis used in the analysis of variance test states that µ1 = µ2 = µ3. ii. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly. iii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. For the population means, the alternate hypothe...

Learning Objective: 4

award:

66.

10

points

.00

i. If the computed value of F is 4.01 and the crit... i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA is µ1 - µ2 = µ3 (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. If the computed value of F is 4.01 and the crit...

Learning Objective: 4

23 of 33

award:

67.

10

points

.00

i. For the population means, the alternate hypothe... i. For the population means, the alternate hypothesis used in the analysis of variance test states that µ1 = µ2 = µ3. ii. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly. iii. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. (i), (ii) and (iii) are all correct statements

→

(ii) is a correct statement, but not (i) or (iii). (iii) is a correct statements but not (i) or (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. For the population means, the alternate hypothe...

Learning Objective: 4

award:

68.

10

points

.00

i. The alternate hypothesis for ANOVA states that ... i. The alternate hypothesis for ANOVA states that not all the means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal.

iii.

The null hypothesis for an ANOVA is µ1 - µ2 = µ3

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. The alternate hypothesis for ANOVA states that ...

Learning Objective: 4

award:

69.

10

points

.00

i. For an ANOVA test, rejection of the null hypoth... i. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA is µ1 - µ2 = µ3

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. For an ANOVA test, rejection of the null hypoth...

Learning Objective: 4

24 of 33

award:

70.

10

points

.00

In a study of low tar cigarettes, five cigarettes ...

In a study of low tar cigarettes, five cigarettes from each of three brands were tested to see if the mean amount of tar per cigarette differs among the brands. i. The F critical value for alpha = 0.05 is 3.89 ii. If F calculated is 4.75, the decision if α = 0.05 is to reject H0 iii. If the calculated F is 4.74, the decision if α = 0.01 is to not reject H0

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

In a study of low tar cigarettes, five cigarettes ...

Learning Objective: 4

award:

71.

10

points

.00

In a study of low tar cigarettes, five cigarettes ...

In a study of low tar cigarettes, five cigarettes from each of three brands were tested to see if the mean amount of tar per cigarette differs among the brands. i. The F critical value for alpha = 0.05 is 3.74 ii. If F calculated is 4.75, the decision if α = 0.05 is to reject H0 iii. If the calculated F is 4.74, the decision if α = 0.01 is to reject H0 (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii).

→

(i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

In a study of low tar cigarettes, five cigarettes ...

Learning Objective: 4

award:

72.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the critical value of F at the 5% level of significance?

19.45 3.00

→

3.35 3.39 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 4

25 of 33

award:

73.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the critical value of F at the 1% level of significance?

99.46

→

5.49 5.39 4.61 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 4

award:

74.

10

points

.00

Given the following Analysis of Variance table for...

Given the following Analysis of Variance table for three treatments each with six observations.

What is the decision?

→

Reject H0--there is a difference in treatments Fail to reject H0--there is a difference in treatments Reject H0--there is a difference in errors Fail to reject H0--there is a difference in errors None of these statements are true

Multiple Choice

Difficulty: Hard

Given the following Analysis of Variance table for...

Learning Objective: 4

award:

75.

10

points

.00

Suppose we select 20 observations from each of fiv...

Suppose we select 20 observations from each of five treatments. The appropriate degrees of freedom are:

→

4 and 15 5 and 20 4 and 19 4 and 20 Not of these are correct.

Multiple Choice

Difficulty: Easy

Suppose we select 20 observations from each of fiv...

Learning Objective: 4

26 of 33

award:

76.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the sum of squares for the treatment?

67.80

→

58.07 149.34 23.47 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

77.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the sum of squares of the error?

→

67.80 58.07 149.34 23.47 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

78.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What are the degrees of freedom for the numerator of the F ratio?

→

2 3 10 27 None of these statements are true

Multiple Choice

Difficulty: Medium

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

27 of 33

award:

79.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What are the degrees of freedom for the denominator?

3 10

→

27 30 None of these statements are true

Multiple Choice

Difficulty: Medium

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

80.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What are the total degrees of freedom?

27 28

→

29 30 None of these statements are true

Multiple Choice

Difficulty: Medium

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

81.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the mean square for treatments?

2.511 2.151 33.9

→

29.035 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

28 of 33

award:

82.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the mean square for error?

→

2.511 2.151 33.9 29.035 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

83.

10

points

.00

A manufacturer of automobile transmissions uses th...

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.

What is the calculated F? 0.086 1.168

→

11.56 13.50 None of these statements are true

Multiple Choice

Difficulty: Hard

A manufacturer of automobile transmissions uses th...

Learning Objective: 5

award:

84.

10

points

.00

Given the following Analysis of Variance table for...

Given the following Analysis of Variance table for three treatments each with six observations.

What are the degrees of freedom for the numerator and denominator?

3 and 18 2 and 17 3 and 15

→

2 and 15 None of these statements are true

Multiple Choice

Difficulty: Medium

Given the following Analysis of Variance table for...

Learning Objective: 5

29 of 33

award:

85.

10

points

.00

Given the following Analysis of Variance table for...

Given the following Analysis of Variance table for three treatments each with six observations.

What is the critical value of F at the 5% level of significance?

3.29 3.68

→

3.59 3.20 None of these statements are true

Multiple Choice

Difficulty: Medium

Given the following Analysis of Variance table for...

Learning Objective: 5

award:

86.

10

points

.00

Given the following Analysis of Variance table for...

Given the following Analysis of Variance table for three treatments each with six observations.

What is the mean square for treatments?

71.2

71.4 →

558 534 None of these statements are true

Multiple Choice

Difficulty: Hard

Given the following Analysis of Variance table for...

Learning Objective: 5

award:

87.

10

points

.00

Given the following Analysis of Variance table for...

Given the following Analysis of Variance table for three treatments each with six observations.

What is the computed value of F? 7.48

→

7.84 8.84 8.48 None of these statements are true

Multiple Choice

Difficulty: Hard

Given the following Analysis of Variance table for...

Learning Objective: 5

30 of 33

award:

88.

10

points

.00

In a study of low tar cigarettes, five cigarettes ...

In a study of low tar cigarettes, five cigarettes from each of three brands were tested to see if the mean amount of tar per cigarette differs among the brands. i. There are 2 degrees of freedom for the numerator. ii. There are 14 degrees of freedom for the denominator. iii. If the sum of squares for the brands is 0.07, the mean square for brands is 0.035. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii).

→

(i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

In a study of low tar cigarettes, five cigarettes ...

Learning Objective: 5

award:

89.

10

points

.00

In a study of low tar cigarettes, five cigarettes ...

In a study of low tar cigarettes, five cigarettes from each of three brands were tested to see if the mean amount of tar per cigarette differs among the brands. i. There are 2 degrees of freedom for the numerator. ii. There are 12 degrees of freedom for the denominator. iii. If the sum of squares for the brands is 0.07, the mean square for brands is 0.035.

→

(i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

In a study of low tar cigarettes, five cigarettes ...

Learning Objective: 5

award:

90.

10

points

.00

The components of an ANOVA table include: The components of an ANOVA table include:

→

Sum of squares total.

Mean Squares.

Degrees of freedom.

All of these components are part of an ANOVA table.

Multiple Choice

Difficulty: Easy

The components of an ANOVA table include:

Learning Objective: 5

award:

91.

10

points

.00

i. If we want to determine which treatment means d... i. If we want to determine which treatment means differ, one method is confidence intervals. ii. If the confidence interval includes 0, there is no difference in the pair of treatment means. iii. If both end points of a confidence interval are of the same sign, it indicates that the treatment means are not different. (i), (ii) and (iii) are all correct statements

→

(i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii). (ii) and (iii) are correct statements but not (i).

All statements are false

Multiple Choice

Difficulty: Hard

i. If we want to determine which treatment means d...

Learning Objective: 6

31 of 33

award:

92.

10

points

.00

i. If we want to determine which treatment means d... i. If we want to determine which treatment means differ, one method is confidence intervals. ii. If the confidence interval includes 0, there is a difference in the pair of treatment means. iii. If both end points of a confidence interval are of the same sign, it indicates that the treatment means are not different. (i), (ii) and (iii) are all correct statements (i) and (ii) are correct statements, but not (iii). (i) and (iii) are correct statements but not (ii).

→

(i) is a correct statement but not (ii) or (iii).

All statements are false

Multiple Choice

Difficulty: Hard

i. If we want to determine which treatment means d...

Learning Objective: 6

award:

93.

10

points

.00

Based on the comparison between the mean annual in...

Based on the comparison between the mean annual incomes for executives with Undergraduate and Master's Degree or more,

→

A confidence interval shows that the mean annual incomes are not significantly different.

The ANOVA results show that the mean annual incomes are significantly different.

A confidence interval shows that the mean annual incomes are significantly different.

The ANOVA results show that the mean annual incomes are not significantly different.

Multiple Choice Based on the comparison between the mean annual in...

Difficulty: Medium Learning Objective: 6

award:

94.

10

points

.00

When comparing the mean annual incomes for executi...

When comparing the mean annual incomes for executives with a High School education or less and Undergraduate Degree, the 95% confidence interval shows an interval of 11.7 to 42.7 for the difference. This result indicates that: There is no significant difference between the two incomes.

The interval contains a difference of zero.

→

Executives with and Undergraduate Degree earn significantly more than executives with a High School education or less.

Executives with and Undergraduate Degree earn significantly less than executives with a High School education or less.

Multiple Choice When comparing the mean annual incomes for executi...

Difficulty: Hard Learning Objective: 6

award:

95.

10

points

.00

In ANOVA analysis, when the null hypothesis is rej...

In ANOVA analysis, when the null hypothesis is rejected, we can find which means are different by:

→

constructing confidence intervals. adding another treatment. doing an additional ANOVA. doing a t test.

None of these statements are true.

Multiple Choice

Difficulty: Medium

In ANOVA analysis, when the null hypothesis is rej...

Learning Objective: 6

32 of 33

award:

96.

10

points

.00

A random sample of 30 executives from companies wi...

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The ANOVA comparing the average income among three levels of education rejected the null hypothesis. The Mean Square Error (MSE) was 243.7. The following table summarized the results:

When comparing the mean salaries to test for differences between treatment means, the t statistic is based on: The treatment degrees of freedom.

The total degrees of freedom.

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The error degrees of freedom The ratio of treatment and error degrees of freedom

Multiple Choice A random sample of 30 executives from companies wi...

Difficulty: Medium Learning Objective: 6

award:

97.

10

points

.00

A random sample of 30 executives from companies wi...

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The ANOVA comparing the average income among three levels of education rejected the null hypothesis. The Mean Square Error (MSE) was 243.7. The following table summarized the results:

When comparing the mean annual incomes for executives with Undergraduate and Master's Degree or more, the following 95% confidence interval can be constructed:

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2.0 ± 2.052*6.51 2.0 ± 3.182*6.51 2.0 ± 2.052*42.46 None of these statements are true

Multiple Choice A random sample of 30 executives from companies wi...

Difficulty: Hard Learning Objective: 6

award:

98.

10

points

.00

If the confidence interval for the difference in t...

If the confidence interval for the difference in treatment means contains zero, then: We can conclude that there is a significant difference in the selected treatment means.

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We can conclude that there is no significant difference in the selected treatment means.

We can conclude that the selected treatment means both equal zero.

We can conclude that at least one of the selected treatment means equals zero.

Multiple Choice

Difficulty: Easy

If the confidence interval for the difference in t...

Learning Objective: 6

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